Koobee Wublee wrote on Sat, 10 May 2008 21:42:09 -0700:

and I do not see any complaint from that multi-year super-senior.

Sorry by not understanding this jargon but what is a "multi-year super-
senior"?

--
http://canonicalscience.org/en/misce...uidelines.html

On May 11, 7:41*am, "Juan R." González-Álvarez
wrote:
Koobee Wublee wrote on Sat, 10 May 2008 21:42:09 -0700:

and I do not see any complaint from that multi-year super-senior.

Sorry by not understanding this jargon but what is a "multi-year super-
senior"?

--http://canonicalscience.org/en/miscellaneouszone/guidelines.html

Senior is the fourth year of college in US, before diploma. Mutli -
year senior is a struggling one to get through, like Erica.

Mike

On May 10, 11:57 pm, JanPB wrote:
On May 10, 9:42 pm, Koobee Wublee wrote:

ds^2 = G c^2 dt^2 / (1 + K^2 / r^2) – 4 r^2 (1 + K^2 / r^2) dr^2 / K^2
– (r^2 + K^2)^2 dO^2 / K^2

Its Ricci and Einstein tensors both vanish. In doing so, it is also
sharing the same coordinate system with the Schwarzschild metric.

Calculate the Ricci tensor. It's not zero. The R_theta,theta is
probably the easiest nonzero component to compute.

What do you mean ‘probably’? Remember dO^2 consists of r, theta, and
phi as the Ricci and the Einstein tensors are only valid in four
dimensions. Since you have not computed the Ricci tensor, you have no
right to call the above solution not a null result of the Ricci and
the Einstein tensors. shrug There is no point to continue.

On May 10, 11:19 pm, Eric Gisse wrote:
On May 10, 8:42 pm, Koobee Wublee wrote:

Its Ricci and Einstein tensors both vanish. In doing so, it is also
sharing the same coordinate system with the Schwarzschild metric.

Then evaluate surface area of a sphere in both coordinate systems.
You'll see something interesting if you actually do the computation.

The multi-year super-senior has nothing to say besides the stupid
question in which I have given the answers already more than once.

There is no short cut in such type of mathematics. shrug

Based on your years of not studying the subject?

Equivalent years, yes. shrug

But mathematically, its Ricci and Einstein tensors also vanish.
shrug

Not if they use the Schwarzschild coordinate system.

Since both metrics use the same coordinate system, you are just
whining about something you do not understand. shrug

Perhaps, there is a flaw in your Cartan whatever. As I said, anyone
who possesses mathematical software like Eric Gisse can easily verify
it, and I do not see any complaint from that multi-year super-senior.
shrug

That's because we [JanPB, me] understand that the r you use is
different from the r in Schwarzschild even if you don't.

Hahaha, is this the best I can do with an amateur tag-team of JanPB
the queer of England also a part-time film critics of second rated
movies and Gisse the multi-year super-senior who sits on his piles of
books for fun?

You appear to think that the coordinates determine the physics.

On the contrary, I have been telling you the exact opposite. shrug
There is no need to continue.

On May 11, 9:05*pm, Koobee Wublee wrote:
On May 10, 11:19 pm, Eric Gisse wrote:

On May 10, 8:42 pm, Koobee Wublee wrote:
Its Ricci and Einstein tensors both vanish. *In doing so, it is also
sharing the same coordinate system with the Schwarzschild metric.

Then evaluate surface area of a sphere in both coordinate systems.
You'll see something interesting if you actually do the computation.

The multi-year super-senior has nothing to say besides the stupid
question in which I have given the answers already more than once.

Then provide the posts where you drive the surface area of spheres of
constant r and t in both supposedly-different metrics.

[snip]

On May 11, 9:53 pm, Koobee Wublee wrote:
On May 10, 11:57 pm, JanPB wrote:

On May 10, 9:42 pm, Koobee Wublee wrote:
ds^2 = G c^2 dt^2 / (1 + K^2 / r^2) – 4 r^2 (1 + K^2 / r^2) dr^2 / K^2
– (r^2 + K^2)^2 dO^2 / K^2

Its Ricci and Einstein tensors both vanish. In doing so, it is also
sharing the same coordinate system with the Schwarzschild metric.

Calculate the Ricci tensor. It's not zero. The R_theta,theta is
probably the easiest nonzero component to compute.

What do you mean ‘probably’?

I meant I computed only R_theta,theta and didn't bother with the rest.

Remember dO^2 consists of r, theta, and
phi as the Ricci and the Einstein tensors are only valid in four
dimensions. Since you have not computed the Ricci tensor, you have no
right to call the above solution not a null result of the Ricci and
the Einstein tensors. shrug There is no point to continue.

If Ricci tensor were zero, then R_theta,theta would be zero. It isn't,
hence Ricci is not zero.

--
Jan Bielawski

On May 11, 11:43 pm, JanPB wrote:
On May 11, 9:53 pm, Koobee Wublee wrote:

ds^2 = G c^2 dt^2 / (1 + K^2 / r^2) – 4 r^2 (1 + K^2 / r^2) dr^2 / K^2
– (r^2 + K^2)^2 dO^2 / K^2

Its Ricci and Einstein tensors both vanish. In doing so, it is also
sharing the same coordinate system with the Schwarzschild metric.

I meant I computed only R_theta,theta and didn't bother with the rest.

Remember dO^2 consists of r, theta, and
phi as the Ricci and the Einstein tensors are only valid in four
dimensions. Since you have not computed the Ricci tensor, you have no
right to call the above solution not a null result of the Ricci and
the Einstein tensors. shrug There is no point to continue.

If Ricci tensor were zero, then R_theta,theta would be zero. It isn't,
hence Ricci is not zero.

Looking for a generic solution that is static and spherically
symmetric, we employ the spherically symmetric coordinate system that
describes a segment of spacetime in general of the following.

ds^2 = c^2 T(r) dt^2 – P(r) dr^2 – Q(r) dO^2

Where

** T(r), P(r), Q(r) = Functions of r only
** dO^2 = cos^2(Phi) dTheta^2 + dPhi^2

Then, the null Einstein tensor (not Ricci) tensor (in free space) is
represented by the following 3 differential equations. You are
supposed to get 4, but 2 of these are identical.

** 1 + dQ/dr dP/dr / (2 P^2) + (dQ/dr)^2 / (4 P Q) – d^2Q/dr^2 / P =
0

** 1 – dQ/dr dT/dr / (1 T P) – (dQ/dr)^2 / (4 P Q) = 0

** - d^2T/dr^2 / T + dT/dr dP/dr / (2 T P) + (dT/dr)^2 / (2 T^2) – dQ/
dr dT/dr / (2 Q T) + dP/dr dQ/dr / (2 P Q) + (dQ/dr)^2 / (2 Q^2) –
d^2Q/dr^2 / Q = 0

There are an infinite number of solutions found. A few examples are
the following in the order of first discovered. These include the
inverse cubed law one at the beginning of the post.

** Schwarzschild’s original solution

ds^2 = G c^2 dt^2 (1 - K / R) – r^4 dr^2 / R^4 / (1 - K / R) – R^2
dO^2

Where

** R = (r^3 + K^3)^(1/3)

** Schwarzschild (Hilbert’s) solution

ds^2 = G c^2 (1 + K / r) dt^2 – dr^2 / (1 + K / r) – r^2 dO^2

** ?’s solution

ds^2 = G c^2 dt^2 / (1 + K / r) – dr^2 (1 + K / r) – (r + K)^2 dO^2

You need to resolve your own mathematical errors. I cannot help you
on that one. shrug

The bottom line is that the Newtonian solution is not unique under the
Einstein field equations, and you need to get over with that. shrug
It is also time to vacate from that fat castle in the air, your
majesty, the queer of England.

On May 13, 1:55*am, Koobee Wublee wrote:
On May 11, 11:43 pm, JanPB wrote:





On May 11, 9:53 pm, Koobee Wublee wrote:
ds^2 = G c^2 dt^2 / (1 + K^2 / r^2) – 4 r^2 (1 + K^2 / r^2) dr^2 / K^2
* * *– (r^2 + K^2)^2 dO^2 / K^2

Its Ricci and Einstein tensors both vanish. *In doing so, it is also
sharing the same coordinate system with the Schwarzschild metric.

I meant I computed only R_theta,theta and didn't bother with the rest.

Remember dO^2 consists of r, theta, and
phi as the Ricci and the Einstein tensors are only valid in four
dimensions. Since you have not computed the Ricci tensor, you have no
right to call the above solution not a null result of the Ricci and
the Einstein tensors. *shrug *There is no point to continue.

If Ricci tensor were zero, then R_theta,theta would be zero. It isn't,
hence Ricci is not zero.

Looking for a generic solution that is static and spherically
symmetric, we employ the spherically symmetric coordinate system that
describes a segment of spacetime in general of the following.

ds^2 = c^2 T(r) dt^2 – P(r) dr^2 – Q(r) dO^2

Where

** *T(r), P(r), Q(r) = Functions of r only
** *dO^2 = cos^2(Phi) dTheta^2 + dPhi^2

Then, the null Einstein tensor (not Ricci) tensor (in free space) is
represented by the following 3 differential equations. *You are
supposed to get 4, but 2 of these are identical.

** *1 + dQ/dr dP/dr / (2 P^2) + (dQ/dr)^2 / (4 P Q) – d^2Q/dr^2 / P =
0

** *1 – dQ/dr dT/dr / (1 T P) – (dQ/dr)^2 / (4 P Q) = 0

** *- d^2T/dr^2 / T + dT/dr dP/dr / (2 T P) + (dT/dr)^2 / (2 T^2) – dQ/
dr dT/dr / (2 Q T) + dP/dr dQ/dr / (2 P Q) + (dQ/dr)^2 / (2 Q^2) –
d^2Q/dr^2 / Q = 0

There are an infinite number of solutions found. *A few examples are
the following in the order of first discovered. *These include the
inverse cubed law one at the beginning of the post.

Of course there are infinite solutions. But they will not accept this.
What they do is select the solution they want to get as a result. Then
work the math to get that solutions. Then claim they got the solution.

Nice, eh?

Mike





** *Schwarzschild’s original solution

ds^2 = G c^2 dt^2 (1 - K / R) – r^4 dr^2 / R^4 / (1 - K / R) – R^2
dO^2

Where

** *R = (r^3 + K^3)^(1/3)

** *Schwarzschild (Hilbert’s) solution

ds^2 = G c^2 (1 + K / r) dt^2 – dr^2 / (1 + K / r) – r^2 dO^2

** *?’s solution

ds^2 = G c^2 dt^2 / (1 + K / r) – dr^2 (1 + K / r) – (r + K)^2 dO^2

You need to resolve your own mathematical errors. *I cannot help you
on that one. *shrug

The bottom line is that the Newtonian solution is not unique under the
Einstein field equations, and you need to get over with that. *shrug
It is also time to vacate from that fat castle in the air, your
majesty, the queer of England.- Hide quoted text -

- Show quoted text -

On May 12, 9:55*pm, Koobee Wublee wrote:

[...]

You continue to say "infinite number" like that's supposed to have any
meaning. There is an infinite number of coordinate systems that the
Schwarzschild solution can be projected on, like any other metric.
What makes me wonder is why you think this is meaningful or even
relevant to the uniqueness of the solution.

You write down multiple representations of the same thing and expect
us to agree with you that they are different. It doesn't work that way.

On May 12, 11:16 pm, Eric Gisse wrote:
On May 12, 9:55 pm, Koobee Wublee wrote:

You continue to say "infinite number" like that's supposed to have any
meaning.

So, you don’t understand what infinity represents.

There is an infinite number of coordinate systems

Yes. shrug

that the Schwarzschild solution can be projected on,

The Schwarzschild metric only applies to the spherically symmetric
polar coordinate system and nothing else. So, you do not understand
the Schwarzschild metric, either.

like any other metric.

Other metric as a solution to the Einstein field equations also only
applies to the spherically symmetric polar coordinate system and
nothing else, and you have never understood the field equations.
shrug

What makes me wonder is why you think this is meaningful or even
relevant to the uniqueness of the solution.

You remain a multi-year super-senior. shrug

You write down multiple representations of the same thing and expect
us to agree with you that they are different. It doesn't work that way.

You were expecting to graduate this spring. Did it happen? No.
shrug